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1. Let D be the random variable with value equal to the number showing on a toss of a
six sided die. Let L be the random variable with value equal to the least number of four
tosses of the die. Describe the probability functions D and L. That is, describe p (D
Solution
When I look at this problem, I see AE between D and r and likewise between L and r and O and
r. If this is some other sign, you can easily correct these answer.
1. P(D = r) = 1/6 for r = 1, 2, 3, 4, 5, 6
L is the minimum of the four tosses.
P(min of 4 tosses) = r = for r = 6, (1/6)^4 = 1/1296
For r <6, it is P(all tosses >= r) - P(all tosses >= r+1) = ((7-r)/6)^4 - ((7-(r+1))/6)^4 = ((7-r)/6)^4
- ((6-r)/6)^4
Note that this formula also works for r=6, so we may simply write
P(L=r) = ((7-r)/6)^4 - ((6-r)/6)^4 for r = 1, 2, 3, 4, 5, 6
We may give the values explicitly
P(1) = 1296/1296 - 625/1296 = 671/1296
P(2) = 625/1296 - 256/1296 = 369/1296 = 41/144
P(3) = 256/1296 - 81/1296 = 175/1296
P(4) = 81/1296 - 16/1296 = 65/1296
P(5) = 16/1296 - 1/1296 = 15/1296 = 5/432
P(6) = 1/1296 - 0/1296 = 1/1296
2.
There are many ways to solve this problem.
One ways is to list all 2^5 = 32 possibilities and count the number with each value of r.
One way, which is a bit more work, is the following.
Clearly, P(O = 0) = P(O = 5) = 1/2^5 = 1/32 (all 0s and all 1s implies 1 way for each).
P(O = 4) = 2/32 = 1/16 (either the first four or last four are all 1s).
P(O=1) = (there are 5 ways of only having 1 1; there are C(5,2) = 10 ways of having 2 1's, but
we remove the 4 with consecutive 1s is 10 - 4 = 6. There is one way to have three 1s not
consecutive, 10101. Thus, the total is 5 + 10 - 4 + 1 = 12
P(O = 1) = 12/32 = 3/8
P(O = 3) either the 3 middle numbers are all 1s, or the first 3 are 1s, the fourth is 0, and the fifth
is easier 0 or 1, or the last three are 1s, the second is 0, and the first is either 0 or 1. 1 + 2 + 2 = 5
P(O = 3) = 5/32
P(O=2) The easiest way to figure this is to simply substract the sum of the other probabilities
from 1. We get 1 - 1/32 - 1/32 - 1/16 - 3/8 - 5/32 = 11/32
We can calculate this a different way, thus checking the other answers.
We have the first two as 1s, the third as a 0, and the fourth and fifth anything (4 ways) + the
fourth and fifth 1s, the third 0, and the first two anything (4 ways) - the first and last two all 1s
and the third 0. Then, we have the first 0, the second and third 1s, the fourth 0, and the fifth 0 or
1 (2 ways), and the third and fourth 1s, the fifth and second 0s, and the first 0 or 1 (2 ways), for a
total of 4+4-1+2+2 = 11
Thus, P(O = 2) = 11/32; this matches our other calculation of it, which gives us faith in our other
answers.
P(O = 0) = 1/32
P(O = 1) = 3/8
P(O = 2) = 11/32
P(O = 3) = 5/32
P(O = 4) = 1/16
P(O = 5) = 1/2^5 = 1/32

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1. Let D be the random variable with value equal to the number showi.pdf

  • 1. 1. Let D be the random variable with value equal to the number showing on a toss of a six sided die. Let L be the random variable with value equal to the least number of four tosses of the die. Describe the probability functions D and L. That is, describe p (D Solution When I look at this problem, I see AE between D and r and likewise between L and r and O and r. If this is some other sign, you can easily correct these answer. 1. P(D = r) = 1/6 for r = 1, 2, 3, 4, 5, 6 L is the minimum of the four tosses. P(min of 4 tosses) = r = for r = 6, (1/6)^4 = 1/1296 For r <6, it is P(all tosses >= r) - P(all tosses >= r+1) = ((7-r)/6)^4 - ((7-(r+1))/6)^4 = ((7-r)/6)^4 - ((6-r)/6)^4 Note that this formula also works for r=6, so we may simply write P(L=r) = ((7-r)/6)^4 - ((6-r)/6)^4 for r = 1, 2, 3, 4, 5, 6 We may give the values explicitly P(1) = 1296/1296 - 625/1296 = 671/1296 P(2) = 625/1296 - 256/1296 = 369/1296 = 41/144 P(3) = 256/1296 - 81/1296 = 175/1296 P(4) = 81/1296 - 16/1296 = 65/1296 P(5) = 16/1296 - 1/1296 = 15/1296 = 5/432 P(6) = 1/1296 - 0/1296 = 1/1296 2. There are many ways to solve this problem. One ways is to list all 2^5 = 32 possibilities and count the number with each value of r.
  • 2. One way, which is a bit more work, is the following. Clearly, P(O = 0) = P(O = 5) = 1/2^5 = 1/32 (all 0s and all 1s implies 1 way for each). P(O = 4) = 2/32 = 1/16 (either the first four or last four are all 1s). P(O=1) = (there are 5 ways of only having 1 1; there are C(5,2) = 10 ways of having 2 1's, but we remove the 4 with consecutive 1s is 10 - 4 = 6. There is one way to have three 1s not consecutive, 10101. Thus, the total is 5 + 10 - 4 + 1 = 12 P(O = 1) = 12/32 = 3/8 P(O = 3) either the 3 middle numbers are all 1s, or the first 3 are 1s, the fourth is 0, and the fifth is easier 0 or 1, or the last three are 1s, the second is 0, and the first is either 0 or 1. 1 + 2 + 2 = 5 P(O = 3) = 5/32 P(O=2) The easiest way to figure this is to simply substract the sum of the other probabilities from 1. We get 1 - 1/32 - 1/32 - 1/16 - 3/8 - 5/32 = 11/32 We can calculate this a different way, thus checking the other answers. We have the first two as 1s, the third as a 0, and the fourth and fifth anything (4 ways) + the fourth and fifth 1s, the third 0, and the first two anything (4 ways) - the first and last two all 1s and the third 0. Then, we have the first 0, the second and third 1s, the fourth 0, and the fifth 0 or 1 (2 ways), and the third and fourth 1s, the fifth and second 0s, and the first 0 or 1 (2 ways), for a total of 4+4-1+2+2 = 11 Thus, P(O = 2) = 11/32; this matches our other calculation of it, which gives us faith in our other answers. P(O = 0) = 1/32 P(O = 1) = 3/8 P(O = 2) = 11/32 P(O = 3) = 5/32 P(O = 4) = 1/16 P(O = 5) = 1/2^5 = 1/32